Question: Find the roots of
\[6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0.\]Enter the roots, separated by commas.
Answer: Dividing the equation by $x^2,$ we get
\[6x^2 - 35x + 62 - \frac{35}{x} + \frac{6}{x^2} = 0.\]Let $y = x + \frac{1}{x}.$  Then
\[y^2 = x^2 + 2 + \frac{1}{x^2},\]so $x^2 + \frac{1}{x^2} = y^2 - 2.$  Thus, we can re-write the equation above as
\[6(y^2 - 2) - 35y + 62 = 0.\]This simplifies to $6y^2 - 35y + 50 = 0.$  The roots are $y = \frac{5}{2}$ and $y = \frac{10}{3}.$

The roots to
\[x + \frac{1}{x} = \frac{5}{2}\]are 2 and $\frac{1}{2}.$  The roots to
\[x + \frac{1}{x} = \frac{10}{3}\]are 3 and $\frac{1}{3}.$

Thus, the roots of $6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0$ are $\boxed{2, 3, \frac{1}{2}, \frac{1}{3}}.$